Determine the boiling point of a solution formed by dissolving 0.187 moles of glucose in 456 grams of water.(For water, K_b = 0.512°C/m)

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:

molality = (moles of solute) / (mass of solvent in kg)
molality = 0.187 mol / (456 g water / 1000 g) = 0.00041 mol/g

Next, we can use the following equation to calculate the boiling point elevation:

ΔTb = molality x Kb x i

where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for water (0.512°C/m), and i is the van't Hoff factor, which is the number of particles into which the solute dissociates (in this case, glucose does not dissociate, so i = 1).

ΔTb = 0.00041 mol/g x 0.512°C/m x 1 = 0.000210°C

Finally, we can calculate the boiling point of the solution:

boiling point = normal boiling point of water + ΔTb
boiling point = 100°C + 0.000210°C = 100.476°C

Therefore, the correct answer is (D) 100.476°C.

Determine the Boiling Point of a Solution Formed by Dissolving